Problem set 9, Harmonic Analysis

Due April 5

Problem 1

1. The set $latex L^1(\mathbb R^d)$ of integrable functions is a complex vector space, and $latex f\mapsto \int f$ is a linear functional.
2. $latex \displaystyle ||f||_1 = \int f$ is a norm on $latex L^1(\mathbb R^d)$, when $latex L^1(\mathbb R^d)$ is seen as a set of equivalence classes of $latex f\sim g$ if and only if $latex f=g$ a.e.

Problem 2

Let $latex L^2(\mathbb R^d)$ be the set of measurable functions $latex f$ such that $latex \displaystyle \int |f|^2 < \infty$, seen as a set of equivalence classes of $latex f\sim g$ if and only if $latex f=g$ a.e.

1. $latex L^2(\mathbb R^d)$ is a complex vector space.
2. The bilinear form $latex \displaystyle \langle f, g \rangle = \int f \bar g$ is well defined on $latex L^2(\mathbb R^d)$ and is an inner product.
3. $latex L^2(\mathbb R^d)$ is complete with the norm $latex \displaystyle ||f||_2 = \sqrt{\langle f, f \rangle}.$ (Hint: Proceed as in the case of $latex L^1$ seen in class.)

Problem 3

1. If $latex f\in L_\text{loc}^1(\mathbb R^d)$, then $latex Mf$ might be infinite at every point.
2. If $latex f\in L^1(\mathbb R^d)$, then $latex Mf$ might be infinite at some points.

Problem 4

We say that $latex \{K_t\}_{t>0}$ is a family of better kernels if it satisfies

• $latex \displaystyle \int_{\mathbb R^d} K_t(x) dx = 1$ for all $latex t>0$;
• there exists $latex A>0$ such that $latex |K_t(x)| \le \dfrac{A}{t^d}$ for all $latex x\in\mathbb R^d$ and $latex t>0$; and 
• there exists $latex A'>0$ such that $latex |K_t(x)| \le \dfrac{A't}{|x|^{d+1}}$ for all $latex x\in\mathbb R^d$ and $latex t>0$.

1. If $latex \Phi\in L^1(\mathbb R^d)$ and $latex |\Phi(x)|\le A/(1 + |x|)^{d+1}$, then its dilations $latex \{\Phi_t\}_{t>0}$ form a family of better kernels.
2. If $latex \{K_t\}_{t>0}$ is a family of better kernels, then there exists a constant $latex c>0$ such that, if $latex f\in L^1(\mathbb R^d)$, $latex |K_t*f(x)| \le cMf(x)$ for all $latex x\in\mathbb R^d$ and $latex t>0$.
3. If $latex \{K_t\}_{t>0}$ is a family of better kernels and $latex f\in L^1(\mathbb R^d)$, then $latex \displaystyle \lim_{t\to0} K_t*f(x) = f(x)$ for almost every $latex x\in\mathbb R^d$.

Problem 5

Let $latex \{K_t\}_{t>0}$ be a family of better kernels and $latex \theta>0$.

1. There exists a constant $latex c_\theta>0$ such that, for $latex f\in L^1(\mathbb R^d)$ and $latex x\in\mathbb R^d$, $latex |K_t*f(y)| \le c_\theta Mf(x)$ for all $latex (y,t)\in \Gamma_\theta(x)$.
2. If $latex f\in L^1(\mathbb R^d)$, then $latex \displaystyle \lim_{\substack{(y,t)\to(x,0)\\(y,t)\in \Gamma_\theta(x)}} K_t*f(y) = f(x)$ for almost every $latex x\in \mathbb R^d$.